P(A|C)≥P(B|C) and P(A|C')≥P(B|C')
Prove that P(A)≥P(B).
First I started with,
P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).
From above,
P(AC)/P(C)≥P(BC)/P(C) and since P(C)≥0,
P(AC)≥P(BC).
Similarly from above,
P(AC')≥P(BC').
Also, P(AC)=P(A)+P(C)-P(AUC) which led to what is below.
As of now I've gotten down to... but got stuck
P(A)≥P(B)-P(BUC)+P(AUC)
P(A)≥P(B)-P(BUC')+P(AUC')